I, for one, simply can't understand the text. For example, what does "at every time the direction of move of $B$ goes from $A$" mean? Could you write this again in other (more intelligible ) words/phrases?

Well excuse me for bad explaining. Well I mean that at each time if person $A$ stops and person $A$ continue its path with its direction at this time it catch $A$. I hope you have understood what I want to say.If you haven't ask me I explain more.

Well as is shown on figure direction of the speed vector of $B$ at each time is aline passing through $A$.Speed of $A$ is $v$ and speed of $B$ is $v'$,and $v \geq v'$. Now we know on what line $A$ moves and where is its start point. Find all points that if $B$ start its move at this point it catches $A$.

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Did you realize what's the question ?

Yes, I understood it the second time you explained it. I just don't think it's easy . Thanks for taking the time to make it clear for me.

No solution!!! Well people think easily this problem is very simple.

First of all, let's find the trajectory of $B$ Well, it's not strictly necessary, but it's nice : let $(x_A(t), y_A(t))$ be the position of $A$ at time $t$ and let, wlog, $x_A(t)=0\ \forall t$ and $y_A(t)=v_At$; similarly let $(x_B(t),y_B(t))$ be the positione of $B$ at time $t$. Let me now use $x$ instead of $x_B(t)$ and $y$ instead of $y_B(t)$. Thanks to the fact that $B$ has constant speed, we can write \[\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2=v_B^2\] or \[\left(\frac{dx}{dt}\right)^2\left[1+\left(\frac{dx}{dy}\right)^2\right]\qquad (1)\] Then, we may state that \[v_At-y=\frac{dy}{dx}x\] because $B$ "follows" $A$. Taking the derivative in t, we get \[v_A-\frac{dy}{dt}=\frac{d^2y}{dx^2}\frac{dx}{dt}x + \frac{dy}{dx}\frac{dx}{dt}\qquad (2)\] Now, getting $\frac{dx}{dt}$ from (2) and substituting in (1) we get \[1+\left(\frac{dy}{dx}\right)^2=\frac{v_B^2}{v_A^2}x^2\left(\frac{d^2y}{dx^2}\right)^2\] Separating the variables we get (let $k=v_B/v_A$): \[\frac{dy}{dx}=\frac{1}{2}\left(Cx^{-1/k}-\frac{1}{C}x^{1/k}\right)\] By knowing that, at $x=x_B(0)$, $\frac{dy}{dx}=y_B(0)/x_B(0)$ we can find C. Integrating again, we get, if $k\neq 1$: \[y=\frac{1}{2}\left[\frac{Ck}{1-k}x^{1-1/k}+\frac{k}{C(1+k)}x^{1+1/k}\right] + D\] where we can know $D$ by imposing the value of $y_B(0)$. Know, for evry choice of $C$ and $D$, if $0<k<1$, this curve has an asymptote for $x=0$, unless, obviously, $x_B(0)=0$ If $k=1$, the equation is more pleasant: \[y=\frac{1}{2}\left(\frac{x^2}{C}-C\log(x)\right) + D\] which, again, has an asymptote in $x=0$, unless, obviously, $x_B(0)=0$. The problem is more difficult if $A$ moves on a curve, instead of a straight line...I don't thinks it's solvable in the most general case...even if $A$ moves along a circle, the calculations grow horribly.

Well I see your solution is very huge.I'll send a shorter solution soon

EvaristeG wrote: First of all, let's find the trajectory of $B$ Well, it's not strictly necessary, but it's nice : This means that mine is not the shortest way...I only thought that it'd be interesting to find this trajectory... A shortest solution can be: let $A$ be in the point $(0,0)$ and move on the straight line $x=0$; 1) if $B$ is, at the beginning, in a point of the halfplane $x\leq 0$ he won't catch $A$ unless he too starts from $(0,0)$; this can be easily seen from the fact that $B$ moves on a curve (not necessarily a straight line) and that $v_B\leq v_A$; 2) if $B$ is in the halfplane $x>0$, we have to find out if it's possible to get to a moment when $A$ and $B$ have the same $y$ and different $x$s; if it is possible, $B$ won't catch $A$ (as we are in the previous case). Let now suppose that, in hypotesis of case 2), $B$ reaches $A$, then the tangent to the trajectory of $B$ will be the line $x=0$ and $y_B\geq y_A$; but if at a given time $y_B\geq y_A$, the tangent to the trajectory of $B$ will be a line with negative slope; since we suppose that $y_B(0)>y_A(0)$, in the moments befor $B$ reaches the line $x=0$ we will have $(y_B-y_A)/(x_B-x_A)\leq 0$, but this would imply that at two given times $t_1>t_0$ we would have $y_A(t_1)<y_A(t_0)$, but this is impossible. (If you draw it and think is easier....it's been terribly hard to write it in a semi-intelligible way)

Only an intuitive sketch. Lets suppose that the prey start from P the hunter from H and it catch the prey at K. The line PK is the trajectory of the prey and the arc (HHK) is the trajectory of the hunter. As Vp > Vh, the angle in K between the two trajectory must be greater or equal to 90 , otherwise is easy to see that the hunter cant catch the prey. As Vp > Vh, |PK| > |(HHK)| >|HK| then it holds that <KHP > <HPK (*). But in some intermediate instant the positions of hunter and prey H and P are s.t. the (*) can be respected, then we have a contradiction. Then the only good positions of the hunter are those on the line (and in front) of the prey.

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Suppose at time $t$, $A$ and $B$ are at points $f(t)=(x_t,y_t), g(t)=(x'_t,y'_t)$. Suppose at each time $y_t=0$ and $x_t$ is increasing now We say that $B$ must be on the line $y=0$ and $x'_0 \geq 0$.Well with this conditions ,it's obvious that $B$ can catch $A$. Now suppose the infinity point $\infty_0=(0,+\infty)$.It is obvious that at time $t$ if the angle $<BA \infty_0$ is bigger than $\frac \pi{2}$ then $B$ can't reach $A$ because $y_t > y'_t$ it is easy to prove this. Now Suppose $h(t)=<BA \infty_0$ at the time $t$. We prove that for every time $t$ that $0<h(t)< \frac \pi{2}$ then there is a time $t'$ that $h(t') \geq 2h(t)$.At time $t$ suppose perpendicular bisector of $AB$ intersect the line $y=0$ at $A'$.Easily $B$ can't catch $A$ till it reaches to point $A'$.And we can prove easily that when $A$ reach at point $A'$ in time $t '$ then $B$ reach to point $B'$ in the triangle $ABA'$ .Now $h(t')=<B'A' \infty_0 \geq <BA' \infty_0 =2<BAA'=2<BA \infty_0=2h(t)$. Well now problem is obvious and eveything is solved.

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sprmnt21 wrote: But in some intermediate instant the positions of hunter and prey H and P are s.t. the (*) can be respected, then we have a contradiction. here is the reason of that: as the tangent h'(t) to the arc (HH'K) varies with continuity, there is a point H' s.t. H'P' is parallel to HK[this result, if I'm not wrong is due to Lagrange]. If <H'KH is less than <HKoo it results that |PK| < |KH'| < arc(H'K) and we are done. If, instead, <H'KH > <HKoo we have <H'KH > 2<HKoo. Then we can apply the same argument to the new position [P', H', K] and son on. But, as the angle in K between PK and the arc(HH'K) is greater than 90 and due to the fact that 2^n <HKoo > 90 for some n, it is sure that we will have <H_nKH less than <HKoo.

Let $ABC$ be an arbitrary triangle. $B$ moves towards $C$ with constant speed size $\alpha$ at each time. $C$ moves towards $A$ with constant speed size $\beta$ at each time. $A$ moves towards $B$ with constant speed size $\gamma$ at each time. i) Determine $\alpha,\beta,\gamma$ dependent on triangle $ABC$ so that triangle $ABC$ be similar to itself at each time. ii) How long does it take for them to meet? (dependent on $a,b,c$) iii) In this case, what is the point $P$ at which $A,B,C$ meet each other? iv) How many times do they go around $P$?

Doesn't any one answer my question? It's a good problem.